习题练习

[{"id":570,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学最喜欢的课外活动调查数据时，制作了如下频数分布表：阅读（12人），运动（18人），音乐（15人），绘画（10人），其他（5人）。如果要将这些数据用扇形统计图表示，那么表示‘运动’这一项的扇形圆心角的度数是多少？","answer":"A","explanation":"首先计算总人数：12 + 18 + 15 + 10 + 5 = 60人。‘运动’所占比例为18 ÷ 60 = 0.3。扇形统计图中整个圆为360度，因此‘运动’对应的圆心角为0.3 × 360 = 108度。故正确答案为A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:46:19","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"108度","is_correct":1},{"id":"B","content":"120度","is_correct":0},{"id":"C","content":"90度","is_correct":0},{"id":"D","content":"72度","is_correct":0}]},{"id":371,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛，共20道题，答对一题得5分，答错或不答扣2分。一名学生最终得分为65分，请问他答对了多少道题？","answer":"A","explanation":"设这名学生答对了x道题，则答错或不答的题数为(20 - x)道。根据题意，答对一题得5分，答错或不答扣2分，总得分为65分，可列出一元一次方程：5x - 2(20 - x) = 65。展开并化简：5x - 40 + 2x = 65，合并同类项得7x - 40 = 65，移项得7x = 105，解得x = 15。因此，该学生答对了15道题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:49:17","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"15","is_correct":1},{"id":"B","content":"14","is_correct":0},{"id":"C","content":"13","is_correct":0},{"id":"D","content":"12","is_correct":0}]},{"id":1800,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某班级组织一次数学知识竞赛，参赛学生的成绩被整理成频数分布表如下：\n\n| 成绩区间（分） | 频数（人） |\n|----------------|------------|\n| 60 ≤ x < 70 | 5 |\n| 70 ≤ x < 80 | 12 |\n| 80 ≤ x < 90 | 18 |\n| 90 ≤ x ≤ 100 | 10 |\n\n已知该班参赛学生总人数为45人，且所有成绩均为整数。若将成绩按从高到低排列，则第23名学生的成绩最可能落在哪个区间？","answer":"C","explanation":"本题考查数据的整理与描述中的频数分布及中位数思想的应用。总人数为45人，将成绩从高到低排列，第23名是正中间的位置，即中位数所在位置。\n\n首先计算累计频数（从高分段开始累加）：\n- 90 ≤ x ≤ 100：10人（第1~10名）\n- 80 ≤ x < 90：18人 → 累计10 + 18 = 28人（第11~28名）\n\n因此，第23名落在第11到第28名之间，即属于“80 ≤ x < 90”这一组。\n\n虽然不能确定具体分数，但根据分组数据的中位数估计方法，第23名最可能落在80到90分区间内。\n\n故正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-06 16:13:28","updated_at":"2026-01-06 16:13:28","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"60 ≤ x < 70","is_correct":0},{"id":"B","content":"70 ≤ x < 80","is_correct":0},{"id":"C","content":"80 ≤ x < 90","is_correct":1},{"id":"D","content":"90 ≤ x ≤ 100","is_correct":0}]},{"id":2428,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究一个实际问题时，构造了一个直角三角形ABC，其中∠C = 90°，AC = 6 cm，BC = 8 cm。他沿斜边AB作了一条高CD，将三角形分为两个小直角三角形ACD和BCD。若该学生进一步测量发现AD的长度为3.6 cm，那么BD的长度应为多少？","answer":"B","explanation":"首先利用勾股定理计算斜边AB的长度：AB = √(AC² + BC²) = √(6² + 8²) = √(36 + 64) = √100 = 10 cm。由于CD是斜边AB上的高，将AB分为AD和BD两段，且AD + BD = AB = 10 cm。已知AD = 3.6 cm，因此BD = 10 - 3.6 = 6.4 cm。此外，也可通过相似三角形验证：△ACD ∽ △ABC，对应边成比例，AC\/AB = AD\/AC → 6\/10 = AD\/6 → AD = 3.6，与题设一致，进一步确认BD = 6.4 cm。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 12:41:01","updated_at":"2026-01-10 12:41:01","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"4.8 cm","is_correct":0},{"id":"B","content":"6.4 cm","is_correct":1},{"id":"C","content":"5.2 cm","is_correct":0},{"id":"D","content":"7.0 cm","is_correct":0}]},{"id":2305,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究轴对称图形时，将一张矩形纸片沿一条直线对折，使得折痕两侧的部分完全重合。已知矩形的长为8 cm，宽为6 cm，若折痕恰好经过矩形的一个顶点和对边上的一点，且该折痕是矩形的对称轴，则这条折痕的长度为多少？","answer":"C","explanation":"本题考查轴对称与勾股定理的综合应用。矩形沿折痕对折后完全重合，说明折痕是图形的对称轴。题目中折痕经过一个顶点和对边上的一点，且为对称轴，意味着折痕是该顶点到对边中点的连线（因为只有这样才能保证对称）。假设矩形ABCD中，A为顶点，对边为CD，则折痕为A到CD中点M的线段AM。在矩形中，AD = 6 cm，DM = 4 cm（因为CD = 8 cm，中点到端点为一半）。在直角三角形ADM中，由勾股定理得：AM² = AD² + DM² = 6² + 4² = 36 + 16 = 52，但此计算错误。正确分析应为：若折痕经过顶点A和对边BC上的点P，且为对称轴，则P应为BC中点。此时AP为折痕。在矩形中，AB = 8 cm，BP = 3 cm（宽的一半），则AP² = AB² + BP² = 8² + 3² = 64 + 9 = 73，故AP = √73 cm。因此正确答案为C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 10:44:46","updated_at":"2026-01-10 10:44:46","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"5 cm","is_correct":0},{"id":"B","content":"√39 cm","is_correct":0},{"id":"C","content":"√73 cm","is_correct":1},{"id":"D","content":"10 cm","is_correct":0}]},{"id":2775,"subject":"历史","grade":"七年级","stage":"初中","type":"选择题","content":"下列哪一项是唐朝对外友好交往的典型事例，体现了当时中外文化交流的繁荣？","answer":"B","explanation":"本题考查唐朝时期中外交流的史实。A项张骞出使西域发生在西汉时期，不属于唐朝；C项郑和下西洋是明朝的事件；D项玄奘西行虽为唐朝中外交流的重要事件，但其主要目的是求取佛经，而鉴真东渡日本则是主动将唐朝的佛教、建筑、医学等文化传播到日本，是唐朝对外友好交往和文化输出的典型代表，更符合‘对外友好交往’和‘文化交流繁荣’的题意。因此，正确答案为B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-12 10:42:55","updated_at":"2026-01-12 10:42:55","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"张骞出使西域，开辟丝绸之路","is_correct":0},{"id":"B","content":"鉴真东渡日本，传播唐朝文化与佛教","is_correct":1},{"id":"C","content":"郑和下西洋，访问亚非多个国家","is_correct":0},{"id":"D","content":"玄奘西行天竺，取回大量佛经并翻译","is_correct":0}]},{"id":2136,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解一元一次方程时，将方程 2(x - 3) = 4 去括号后得到 2x - 6 = 4，然后他\/她接下来应该进行的正确步骤是：","answer":"D","explanation":"方程 2x - 6 = 4 中，-6 是常数项，为了将含 x 的项单独留在左边，应使用等式的基本性质：两边同时加上6，得到 2x = 10。这是解一元一次方程的标准步骤，符合七年级学生对方程解法的学习要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 13:00:46","updated_at":"2026-01-09 13:00:46","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"两边同时加上6","is_correct":0},{"id":"B","content":"两边同时除以2","is_correct":0},{"id":"C","content":"两边同时减去6","is_correct":0},{"id":"D","content":"两边同时加上6","is_correct":1}]},{"id":1006,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某班级组织植树活动，若每名学生种3棵树，则还剩10棵树没人种；若每名学生种4棵树，则最后一名学生只需种2棵树。这个班级共有___名学生。","answer":"12","explanation":"设这个班级共有x名学生。根据题意，树的总数不变。第一种情况：每名学生种3棵，还剩10棵，所以总树数为3x + 10。第二种情况：前(x - 1)名学生每人种4棵，最后一名学生种2棵，总树数为4(x - 1) + 2 = 4x - 4 + 2 = 4x - 2。因为树的数量相同，列方程：3x + 10 = 4x - 2。解这个一元一次方程：移项得10 + 2 = 4x - 3x，即12 = x。所以这个班级共有12名学生。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 05:03:53","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":2197,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在练习本上记录了一周内每天的温度变化情况，规定比前一天升高记为正，降低记为负。已知周一到周二的温度变化为 -3℃，周三到周四的温度变化为 +5℃，周五到周六的温度变化为 -2℃。如果周一的起始温度为 10℃，那么周六的温度是多少？","answer":"B","explanation":"从周一的 10℃ 开始，周二变化 -3℃，温度为 10 - 3 = 7℃；周三到周四变化 +5℃，即温度上升 5℃，变为 7 + 5 = 12℃；周五到周六变化 -2℃，即下降 2℃，变为 12 - 2 = 10℃。因此周六的温度是 10℃，正确答案是 B。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:25:31","updated_at":"2026-01-09 14:25:31","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"8℃","is_correct":0},{"id":"B","content":"10℃","is_correct":1},{"id":"C","content":"12℃","is_correct":0},{"id":"D","content":"14℃","is_correct":0}]},{"id":1960,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在研究某城市一周内的空气质量指数（AQI）变化时，记录了连续7天的AQI数据：45, 68, 52, 73, 60, 55, 80。为了分析这组数据的集中趋势，该学生计算了这组数据的中位数。请问这组AQI数据的中位数是多少？","answer":"B","explanation":"本题考查数据的收集、整理与描述中中位数的概念与计算。中位数是一组数据按从小到大（或从大到小）排列后，处于中间位置的数。首先将AQI数据从小到大排序：45, 52, 55, 60, 68, 73, 80。由于共有7个数据（奇数个），中位数就是第4个数，即60。因此，这组数据的中位数是60。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:47:21","updated_at":"2026-01-07 14:47:21","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"55","is_correct":0},{"id":"B","content":"60","is_correct":1},{"id":"C","content":"68","is_correct":0},{"id":"D","content":"73","is_correct":0}]}]